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Electric Field due to a Charged Cylinder 
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#1
Sep3007, 09:45 AM

P: 171

1. The problem statement, all variables and given/known data
Our professor gave this as sort of a challenge to find on our own; I've been stumped for a while and just can't seem to figure it out. Find the electric field at a point [itex]{P}[/itex] a distance [itex]{Z_{0}}[/itex] along the zaxis due to a charged cylinder of radius [itex]{R_{0}}[/itex], length [itex]{L}[/itex], and uniformly distributed charge [itex]{q}[/itex]. Find the electric field for a (a) Hollow Cylinder and for a (b) Solid Cylinder Yea,...my MS Paint skills could be better. 2. Relevant equations Electric Field Equation [tex] {\vec{E}_{P1}} = {\frac{{k_{e}}{q_{1}}}{{\left(r_{_{1P}}\right)}^{2}}{\hat{r}_{1P}} [/tex] [tex] {{E}_{P1}} = \frac{{k_{e}}{q_{1}}}{{\left(r_{_{1P}}\right)}^{2}} [/tex] [tex] {{dE}_{P1}} = \frac{{k_{e}}{dq}}{{\left(r_{_{1P}}\right)}^{2}} [/tex] Charge Density Equations [tex] {\lambda} = \frac{q}{L} [/tex] [tex] {\sigma} = \frac{q}{A} [/tex] [tex] {\rho_{q}} = \frac{q}{V} [/tex] Electric Field due to a Charged Ring (C.R.) [tex] {E_{P1_{z}}} = \frac{{k_{e}}{q}{Z_{0}}}{\left({{R_{0}}^{2}}+{{Z_{0}}^{2}}\right)^{3/2}} [/tex] Electric Field due to a Charged Disk (C.D.) [tex] {E_{P1_{z}}} = \frac{{2}{k_{e}}{q}{Z_{0}}}{{{R}_{0}}^{2}}{\left[{1}{\frac{{{Z}_{0}}}{\sqrt{\left({{R_{0}}^{2}}+{{Z_{0}}^{2}}\right)}}}\rig ht]} [/tex] 3. The attempt at a solution For easier notation let, [tex] {q} = {q_{1}} [/tex] (a) Hollow Cylinder For the hollow charged cylinder, I treat the cylinder as the sum of rings of differential length [itex]{ds}[/itex] and charge [itex]{dq}[/itex]. That is, a differential segment of the ring has a length [itex]{ds}[/itex] and charge [itex]{dq}[/itex]. I recognize that, [tex] {{dE}_{net}} = {dE}_{P1_{z}} [/tex] Where, [tex] {dE}_{P1_{z}} = {dE}{{cos}{\beta}} [/tex] Now, because I am using a ring I refer to [itex]{\lambda}[/itex], so that, [tex] {\lambda} = {\frac{dq}{ds}} [/tex] However, I can already recognize that this approach is folly because I have already indicated (in the figure) [itex]{ds}[/itex] as a differential width of the cylinder as opposed to a differential length segment. So already I am stuck, any ideas? Particularly, I know how I want to break up the hollow cylinder (in to a sum of differential rings) however, I am not sure how to put it together. Specifically in terms of what constant: [itex]{\lambda}[/itex], [itex]{\sigma}[/itex], or [itex]{{\rho}_{q}}[/itex]? So, that I only have to integrate once to find the electric field. Any help is appreciated. Thanks, PFStudent 


#2
Sep3007, 10:08 AM

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P: 41,482




#3
Sep3007, 10:41 AM

P: 171

However, if I just want to find the electric field (due to the charged cylinder) in one integral will I not have to resort to using [itex]{{\rho}_{q}}[/itex] to set up the differential piece of which I will integrate to get the whole electric field contributed by the charged cylinder in one integral? Basically, I have to resort to using a differential hollow disk of width [itex]{ds}[/itex] and integrate that using [itex]{{\rho}_{q}}[/itex]. Or am I missing something here. To me it just does not make sense how using a differential length segment [itex]{ds}[/itex] of a differential ring will get you the electric field, I mean when you integrate that using [itex]{\lambda}[/itex] will you not just get the elctric field contributed by just that whole ring? As opposed to the whole cylinder...? Yea,...I feel like I am missing something here. Thanks for the reply Doc Al. Any help is appreciated. Thanks, PFStudent 


#4
Sep3007, 11:19 AM

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P: 41,482

Electric Field due to a Charged Cylinder



#5
Sep3007, 11:33 AM

P: 171

What I am driving at is how do you find the electric field due to a hollow charged cylinder in one integral. In particular to find the electric field due to a charged cylinder in one integral how should and what (shape) should I break up my integral in to, so that when I integrate (only once) I will be able to find the electric field due to the charged cylinder. That is where I am stumped. Thanks for the reply. Thanks, PFStudent 


#6
Sep3007, 11:39 AM

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P: 41,482




#7
Sep3007, 01:08 PM

P: 171

[tex] {{dE}_{net}} = {{dE}_{P1_{z}}} [/tex] [tex] {{dE}_{P1_{z}}} = {{dE}_{P1}}{{cos}{\beta}} [/tex] [tex] {{dE}_{P1}} = {\frac{{{k}_{e}}{dq}}{{\left({r}_{{}_{1P}}\right)}^{2}}} [/tex] [tex] {cos}{\beta} \equiv \frac{adj.}{hyp.} [/tex] So, [tex] {cos}{\beta} = {\frac{s}{{\sqrt{{{s}^{2}}+{{{R}_{0}}^{2}}}}}} [/tex] [tex] {{dE}_{P1_{z}}} = {{dE}_{P1}}{{cos}{\beta}} [/tex] [tex] {{dE}_{P1_{z}}} = {\left(\frac{{k_{e}}{dq}}{{\left(r_{_{1P}}\right)}^{2}}\right)}{{\left( \frac{s}{{\sqrt{{{s}^{2}}+{{{R}_{0}}^{2}}}}}\right)}} [/tex] Also, [tex] {\lambda} = {\frac{dq}{ds}} [/tex] So, [tex] {dq} = {\lambda}{ds} [/tex] In addition, [tex] {r_{_{1P}}} = \sqrt{{s}^{2}+{R_{0}}^{2}} [/tex] Ok, I am up to here, however I just want to make sure I understand the integration correctly. For the charged ring a differential (lengthwise) segment of the ring was the [itex]{ds}[/itex]. However for this charged cylinder we are letting a differential width (thickness) of the ring be [itex]{ds}[/itex]. Will the integral still work anyway since we are using [itex]{\lambda}[/itex], but noting how [itex]{ds}[/itex] does not represent the same differential shape (in both the situations above)? I get the feeling as if I should be using [itex]{\sigma}[/itex] (because we are dealing with width (thickness))... So, why does this integration work anyway with [itex]{\lambda}[/itex] as opposed to [itex]{\sigma}[/itex] in one integral? Thanks Doc Al. Thanks, PFStudent 


#8
Sep3007, 01:20 PM

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P: 41,482




#9
Sep3007, 01:56 PM

P: 171

So, even though a cylinder has volume and surface area, that is not the point of focus for the integration. What matters is what we are integrating over, so in this case because we are integrating over a variable length, [itex]{s}[/itex] with a differential length element, [itex]{ds}[/itex] we then use, [itex]{\lambda}[/itex] rather than, [itex]{\sigma}[/itex]. Is that right? So, then because we are integrating over a length we have the charge density of the hollow charged cylinder given as, [tex] {\lambda} = {\frac{q}{L}} [/tex] Where, because the charge is uniformly distributed then [itex]{\lambda}[/itex] must be constant such that for a differential (length) element [itex]{ds}[/itex] with a differential charge [itex]{dq}[/itex]; their ratio must be the same, that is [itex]{\lambda}[/itex]. In other words, [tex] {\lambda} = {\frac{dq}{ds}} [/tex] So that, [tex] {\frac{dq}{ds}} = {\frac{q}{L}} [/tex] Is all that right? Thanks so much for the insights Doc Al. Thanks, PFStudent 


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